3y 1 + 5y 2. y 1 + y 2 20 y 1 0, y 2 0.

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1 1 Linear Programming A linear programming problem is the problem of maximizing (or minimizing) a linear function subject to linear constraints. The constraints may be equalities or inequalities. 1.1 Example Problem. There are two ports P 1 and P 2 where s 1 = 10 and s 2 = 30 tons of some commodity are stored respectively. The demand of the supermarket M in this commodity is minimum 20 tons. The transportation of 1 ton from P 1 to M costs $3 and from P 2 to M costs $5. The problem is to meet the market requirements at minimum transportation cost. Let the amount of commodity shipped from P 1 and P 2 to M be y 1 and y 2 respectively. Then we must minimize the objective function subject of the following constraints 3y 1 + 5y 2 y 1 + y 2 20 y 1 10, y 2 30, y 1 0, y 2 0. Using Maple >with(optimization): > LPSolve(3x + 5y, {x + y >= 20, x <= 10, y <= 30, x >= 0, y >= 0}); [80, [x = 10, y = 10]] 1.2 Graphical Method To solve this problem first we must find the feasible set, that is the set of all points (x, y) R 2 which satisfy all above constraints. For this we must graph 1

2 the solution of each of these inequalities separately and the final feasible set will be the intersection of all these solution sets. This might be: (a) Empty set, in this case the problem is called unfeasible and has no solution; (b) A bounded polygon. In this case the problem is feasible. To solve the problem compute the coordinates of all corner points of the feasible polygon, substitute the coordinates of the corner points into the objective function to see which gives the optimal value. (c) An unbounded polygon. In this case a solution may or may not exist How to graph the solution of a linear inequality Solution of the inequality 3x 4y 12: Step 1. First sketch the line 3x 4y = 12. Find the y-intercept: x = 0 and y can be solved from 3 0 4y = 12, y = 3, so the y-intercept is the point (0, 3). Find the x-intercept: y = 0 and x can be solved from 3 x 4 0 = 12, x = 4, so the x-intercept is (4, 0). This two points are enough to sketch the graph of 3x 4y = 12. This graph divides the plane into two regions, the upper one and the lower one. Which one is the solution of our inequality 3x 4y 12? Step 2. Choose the origin (0, 0) as the test point (since it is not on the line). Substituting x = 0, y = 0 in the inequality gives Since this is a true statement, (0, 0) is in the solution set, so the solution set consists of all points on the same side as (0, 0). Choose some another test point if (0, 0) is on the line. 2

3 1.2.2 How to find corner points To find the corner points we must just solve the systems of linear equations. For example the corner point which corresponds to two constrains ax + by c and px + qy r is the solution of the system { ax + by = c px + qy = r Standard Form of Linear Programming Problem Maximize a linear function (called objective function) subject if following constraints: f(x 1,..., x n ) = c 1 x c n x n a 11 x a 1n x n b 1... a m1 x a mn x n b m x 1 0,... x n 0. In matrix form: Maximize c T x subject of Ax b, x 0. Remark 1. An equality constraint a 11 x a 1n x n = b 1 can be converted to inequality by introducing two inequality constraints a 11 x a 1n x n b 1, a 11 x a 1n x n b 1 or, equivalently, a 11 x a 1n x n b 1, a 11 x... a 1n x n b 1. 3

4 So we can assume that all constraints are given inequalities. Remark 2. Sometimes it is more convenient to work with equality constraints. An inequality constraint a 11 x a 1n x n b can be converted to an equality constraint by adding a new nonnegative variable, w, which is called a slack variable, a 11 x a 1n x n + w = b, w 0. So we can assume that all constraints are given equalities Fundamental Theorem of Linear Programming Each of the constraints of a standard LP problem defines a closed half-space of R n, which is a convex set. The whole feasible region, as the intersection of convex sets, is a convex set too. Given a convex set S, point x 0 is an extreme point, if each line segment that lies completely in S and contains point x 0, has x 0 as an end point of the line segment. Theorem 1 If a LP problem has a solution then there exists an extreme point of the feasible region which is optimal The Transportation Problem There are I ports, or production plants, P 1,..., P I, that supply a certain commodity, and there are J markets, M 1,..., M J, to which this commodity must be shipped. Port P i possesses an amount s i of the commodity (i = 1, 2,..., I), and market M j must receive the amount r j of the commodity (j = 1,..., J). Let b ij be the cost of transporting one unit of the commodity from port P i to market M j. The problem is to meet the market requirements at minimum transportation cost. Let y ij be the quantity of the commodity shipped from port P i to market M j. The total transportation cost is I J b ij y ij. i=1 j=1 4

5 The amount sent from port P i is J j=1 y ij and since the amount available at port P i is s i, we must have J y ij s i, i = 1, 2,..., I. j=1 The amount sent to market M j is I i=1 y ij, and since the amount required there is r j, we must have I y ij r j, j = 1, 2,..., J. i=1 It is assumed that we cannot send a negative amount from P i to M j, we have y ij 0, for i = 1,..., I and j = 1,..., J. Our problem is: minimize I J b ij y ij. i=1 j=1 subject to J y ij s i, i = 1, 2,..., I. j=1 I y ij r i, j = 1, 2,..., J. i=1 y ij 0, i = 1,..., I, j = 1,..., J. 5

6 Exercises Solve the following LP problems using graphical method and check the solution using Maple. 1. Find x, y to minimize x + y subject of the constraints x + 2y 3 2x + y 5 y Find x 1, x 2 to maximize x 1 + x 2 subject of the constraints x 1 + 2x 2 4 4x 1 + 2x 2 12 x 1 + x 2 1 x 1 0, x Consider the LP problem below. The CVH Company of Leeds is a revenue maximizing firm. It has chosen to focus on the production and sale of two goods: A and B. Good A requires 1 hour of skilled labor and 4 kilos of metal. It is a standard product and retails for $40. Good B is a higher quality version of good A. It needs 2 hours of skilled labor and demands 3 kilos of metal. It sells for a slightly higher price of $50 per unit. Under current operating conditions, the firm is prepared to hire 40 hours worth of labor each day and it has a long term contract with a supplier who can deliver 120 kilos of metal each day. Your task is to construct a simple LP and graph to solve. 4. MFG is small car manufacturer producing two models of handmade cars. These cars are prestige vehicles and command a very high price. The two models are well known throughout the specialist car market as X and Y. Both car X and car Y are very profitable. Car X produces $2,000 of profit for each vehicle produced whilst Car Y produces $3,000. the firm is motivated purely by profit. The company cannot produce as many cars as it would like. It cannot produce more than 15 cars in a year because they all have to be hand finished. For operational reasons, there are also limitations on the 6

7 combinations of cars X and Y which can be manufactured in a given year. It is not possible given the current technology for more than 3 model Y cars to be produced for each model X car. This can create a headache for MFG. (a) Write in words MFGs objective function (b) Translate the objective function into a simple expression using X to show the number of model X and Y to denote the number of model Y vehicles that can be manufactured in a year. (c) How many constraints are there to this problem? Work out what the expressions are for these constraints. (d) Construct the linear program. How many of each should be produced? 7